Differentiability

f(x) = x sin(1/x) for every x not equal to 0
0 for x = 0

How do i show whether this function is differentiable or not at x=0. First of all i know that if a function is differentiable at a then lim(h->0) (f(a+h)-f(a))/h has to exist or rather lim(x->a) (f(x)-f(a))/(x-a). I was pretty sure that the function was not differentiable at x=0 until i plugged it into my graphics calc to have a look at the graph, need an explanation for this please.

nobody wanna help me? all i know that if a function is differentiable at point a then the limit lim(h->0) (f(a+h)-f(a))/h exists but i tried doing this with some random function that is obviously not differentiable at some point and yet i managed to prove that it is differentiable which obviously means that i have done something wrong. so can somebody please give me a hand showing f(x) from the previous post (piecewise function btw) is diff or not diff at x=0. any help is appreciated.

nobody wanna help me? all i know that if a function is differentiable at point a then the limit lim(h->0) (f(a+h)-f(a))/h exists but i tried doing this with some random function that is obviously not differentiable at some point and yet i managed to prove that it is differentiable which obviously means that i have done something wrong. so can somebody please give me a hand showing f(x) from the previous post (piecewise function btw) is diff or not diff at x=0. any help is appreciated.

well f is certainly differentiable because the limit is equal to 0. but the problem is, when i use the same logic with g(x) i also get a limit which is equal to 1 but i know that it is not differentiable at x=0. thats what is confusing for me.

EDIT: i used the lim(x->a) f(x)-f(a)/x-a method for finding the limit. doubt it matters though

btw is it right for me to be taking lim(h->0) h sin (1/h) to be zero coz the logic that im using is that -1<sin(k)<1 and h gets really really small hence it will converge to 0 right but it will be an oscillating convergence.

btw is it right for me to be taking lim(h->0) h sin (1/h) to be zero coz the logic that im using is that -1<sin(k)<1 and h gets really really small hence it will converge to 0 right but it will be an oscillating convergence.

Yes, and if you want a formal argument for that intuition you wrote try sandwiching between h and -h. Since both h and -h goes to 0 as h goes to 0, we are through.

Originally Posted by ah-bee

well f is certainly differentiable because the limit is equal to 0. but the problem is, when i use the same logic with g(x) i also get a limit which is equal to 1 but i know that it is not differentiable at x=0. thats what is confusing for me.

EDIT: i used the lim(x->a) f(x)-f(a)/x-a method for finding the limit. doubt it matters though

edit: sorry about the messiness, would be much more neater if i knew how to use the notation stuff that u are using. not sure if it is just an algebraic mistake or something which im pretty sure it isnt, i just did some manipulation using the limit laws.